If |G/H|=4 then G is union of three proper subgroups

Question

If |G/H|=4 and G/H is not cyclic then G is union of three proper subgroups. I understand that there are three subgroups of order two in the G/H group. Consequently, in the group G there are three subgroups A, B, C of index two which in them pass under the homomorphism. But how to prove that G is a union of the subgroups A, B, and C

Answer 1

Hint: If $A,B,C$ are subsets in $G$, then $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.$$

$G/H\cong K_4$, so you know that by the Correspondence theorem that there exists subgroups $A,B,C$ in $G$ such that $|A|=|B|=|C|=2|H|$ and $|A\cap B|=|B\cap C|=|C\cap A|=|A\cap B\cap C|=|H|$, so what can you say about $|A\cup B\cup C|$?